Proof: $ { \mathbb{Q}[x] } / (x^2 +3x +1) \cong \mathbb{Q}(\sqrt{5}) $

Question

I have a problem with proving this: $$ \mathbb{Q}[x] / (x^2 +3x +1) \cong \mathbb{Q}(\sqrt{5}) $$ My observations:

In $$\mathbb{Q}[x] / (x^2 +3x +1)$$ $$x^2 \equiv -1 -3x$$ so $$\forall f \in \mathbb{Q}[x]/(x^2 +3x +1) \qquad(\deg(f) \le 1).$$

Also, we can see that $$x^2 + 3x +1 = (x+\frac 32 +\sqrt5)(x+\frac 32-\sqrt5)$$

And then I got stuck.

Answer 1

A root of $x^2+3x+1$ is $(\sqrt{5}-3)/2$, so $$ \mathbb{Q}[x]\big/(x^2+3x+1)\cong\mathbb{Q}\bigl((\sqrt{5}-3)/2\bigr) $$ Observe that $$ \sqrt{5}=3+2\frac{\sqrt{5}-3}{2} $$ and you're done.

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In general, if $f\in F[x]$ is an irreducible polynomial over the field $F$ and $E$ is an extension field of $F$ such that $f(\alpha)=0$, for some $\alpha\in K$, then $$ F[x]/(f)\cong F(\alpha) $$ Consider the homomorphism $\varphi\colon F[x]\to K$ which is the identity on $F$ and sends $x$ to $\alpha$. Then the image of $\varphi$ is $F[\alpha]$, the smallest subring of $K$ containing $F$ and $\alpha$.

Show that $\ker\varphi=(f)$ and this will give $$ F[x]/(f)\cong F[\alpha] $$ On the other hand, $(f)$ is a maximal ideal in $F[x]$, so $F[\alpha]$ is a field and therefore $F[\alpha]=F(\alpha)$, the smallest subfield of $K$ containing $F$ and $\alpha$.